$3.(25$ points$)$ The following problem is taken from astronomy. Suppose we have two space objects with mass $1$ and $$ in circular rotation in the space. We can think of these two space objects as the earth and the moon, where the moon moves around the earth with distance $1.($This is a normalized distance.$)$ A third object, which is relatively much smaller than the first two, such that it would not make any change in the orbits of the first two, is also moving in the space. You can think of this as a space$-$ship. The equations below describes the movement of the third object, i$.$e$.$ the spaceship. We consider a two space dimensional case. Let the position of the earth be the origin of our coordinate. The solution $($y$1($t$),$ y$2($t$))$ would give the position of the third object. The equations are given as:

y $001=$ y$1+2$y $02$ y$1+$ D$1$ y$1$ D$2$ y $002=$ y$22$y $01$ y$2$ D$1$ y$2$ D$2$ D$1=(($y$1+)2+$ y $22)3/2$ D$2=(($y$1$ $)+$y $22)3/2$ Use parameter: $=0\mathrm{.}012277471,$ $=1,$ y$1(0)=0\mathrm{.}994,$ y $01(0)=0,$ y$2(0)=0,$ y$02(0)=2\mathrm{.}0015851063790825.$ It is known that the solutions would become periodic, i$.$e$.,$ the object will following a certain orbit. When t $=17\mathrm{.}065211656$ it would complete one full round of its orbit. $1.$ Write a program that would solve the system with the classical $4$th order Runge$-$Kutta method, but reduce the step number to $6000.$